Det regulære språket $\{\ \str{aca}, \str{acbb}, \str{bbca}, \str{bbcbb} \}$ over alfabetet $A = \{ \str{a}, \str{b}, \str{c} \}$ er lik
$\{ \str{a}, \str{bb}, \str{c} \}^{\ast} = \{ \Lambda, \str{a}, \str{bb}, \str{c}, \str{aa}, \str{abb}, \str{ac}, \str{bba}, \str{bbbb}, \str{bbc}, \str{ca}, \str{cbb}, \str{cc}, ... \}$ er en uendelig mengde som inneholder den tomme strengen, og kan altså ikke være lik $\{\ \str{aca}, \str{acbb}, \str{bbca}, \str{bbcbb} \}$.
$\{ \str{ac} \} \{ \str{a}, \str{bb} \} = \{ \str{aca}, \str{acbb} \}$
$\{ \str{a}, \str{bb} \} \{ \str{c} \} \{ \str{a}, \str{bb} \} = \{ \str{aca}, \str{acbb}, \str{bbca}, \str{bbcbb} \}$
$\{ \str{a} \} \{ \str{bb}, \str{c} \} \{ \str{a} \} = \{ \str{abba}, \str{aca} \}$
Vi kan bygge opp det regulære språket $\{\ \str{aca}, \str{acbb}, \str{bbca}, \str{bbcbb} \}$ over alfabetet $A$ fra elementene i
basismengden for regulære språk over $A$ slik:
$( \{ \str{a} \} \cup \{ \str{b} \} \{ \str{b} \} ) \{ \str{c} \} ( \{ \str{a} \} \cup \{ \str{b} \} \{ \str{b} \} )$
Det regulære språket $\{\ \str{ab}(\str{bb})^{n}\str{c} \mid n \in \mathbb{N} \} = \{ \str{abc}, \str{abbbc},\str{abbbbbc}, ... \}$ over alfabetet $\{ \str{a}, \str{b}, \str{c} \}$ er lik
$\{ \str{a} \} \{ \str{b} \}^\ast \{ \str{c} \} = \{ \str{ac}, \str{abc}, \str{abbc}, ... \} = \{ \str{a}\str{b}^n\str{c} \mid n \in \mathbb{N} \}$
$\{ \str{a} \} \{ \str{bb} \}^\ast \{ \str{c} \} = \{ \str{ac}, \str{abbc}, \str{abbbbc}, ... \} = \{ \str{a}(\str{bb})^n\str{c} \mid n \in \mathbb{N} \}$
$\{ \str{ab} \} \{ \str{bb} \}^\ast \{ \str{c} \} = \{ \str{abc}, \str{abbbc}, \str{abbbbbc}, ... \} = \{ \str{ab}(\str{bb})^n\str{c} \mid n \in \mathbb{N} \}$
Hvilket av disse regulære språkene er lik språket $L = \set{\str{11}, \str{110}, \str{111}, \str{1101}, \str{11011}, \str{1111}, ...}$?
Strengene i $L$ starter alltid med $\str{11}$, deretter kommer enten $\str{0}$ og et vilkårlig antall $1$-ere, eller bare et vilkårlig antall $1$-ere uten noe $\str{0}$ foran. Til sammen blir dette $\set{\str{11}} \set{\tomstreng, \str{0}} \set{\str{1}}^*$.